6 MECHANICS OF MATERIALS KAYMA GERİLMESİ DAĞILIMI Fifth Edition MECHANICS OF CHAPTER MATERIALS 6 Ferdinand P. Beer E. Russell Johnston, Jr. John T. DeWolf David F. Mazurek EĞİLMEYE MARUZ KİRİŞLERDE Lecture Notes: J. Walt Oler Texas Tech University KAYMA GERİLMESİ DAĞILIMI © 2009 The McGraw-Hill Companies, Inc. All rights reserved.

MECHANICS OF MATERIALS Edition Fifth MECHANICS OF MATERIALS Beer • Johnston • DeWolf • Mazurek Shear on the Horizontal Face of a Beam Element Yandaki kirişte, CDC’D’ elemanının dengesi:  Fx  0  H    D   C dA A  MD  MC H  y dA A I Q   ydA • A Statik Moment M  M  dM x  V x D C dx Buradan, H  VQ x I q  H  VQ  kayma akıkı x I © 2009 The McGraw-Hill Companies, Inc. All rights reserved.

MECHANICS OF MATERIALS Edition Fifth MECHANICS OF MATERIALS Beer • Johnston • DeWolf • Mazurek Shear on the Horizontal Face of a Beam Element Kayma akısı, q  H  VQ x I burada Q   y dA A  y1üstündeki alanıl statik atalet momenti I  I   y2dA z A A'  tümalan ıl atalet momenti Aynı sonuç diğer kısım içinde bulunur q  H   VQ  q x Q  Q  0 H   H I © 2009 The McGraw-Hill Companies, Inc. All rights reserved.

 VQ   It  H  q x  VQ x MECHANICS OF MATERIALS Edition Fifth MECHANICS OF MATERIALS Beer • Johnston • DeWolf • Mazurek Determination of the Shearing Stress in a Beam Ortalama Kayma Gerilmesi  H  q x  VQ x A A I t x  ort  VQ   ort It © 2009 The McGraw-Hill Companies, Inc. All rights reserved.

MECHANICS OF MATERIALS Edition Fifth MECHANICS OF MATERIALS Beer • Johnston • DeWolf • Mazurek Örnek Çözüm © 2009 The McGraw-Hill Companies, Inc. All rights reserved.

MECHANICS OF MATERIALS Edition Fifth MECHANICS OF MATERIALS Beer • Johnston • DeWolf • Mazurek Örnek-devam © 2009 The McGraw-Hill Companies, Inc. All rights reserved.

MECHANICS OF MATERIALS Edition Fifth MECHANICS OF MATERIALS Beer • Johnston • DeWolf • Mazurek Örnek-devam •Tüm kesitin ağırlık merkezi ve atalet momenti © 2009 The McGraw-Hill Companies, Inc. All rights reserved.

MECHANICS OF MATERIALS Edition Fifth MECHANICS OF MATERIALS Beer • Johnston • DeWolf • Mazurek Örnek-devam © 2009 The McGraw-Hill Companies, Inc. All rights reserved.

MECHANICS OF MATERIALS Edition Fifth MECHANICS OF MATERIALS Beer • Johnston • DeWolf • Mazurek Örnek-devam © 2009 The McGraw-Hill Companies, Inc. All rights reserved.

MECHANICS OF MATERIALS Edition Fifth MECHANICS OF MATERIALS Beer • Johnston • DeWolf • Mazurek Örnek-devam   VQ It a max VQc c  , c max Itc min tc min  75mm  VQG VQc   c min tc max  , max Itc max ItG  225mm, tG  75mm © 2009 The McGraw-Hill Companies, Inc. All rights reserved. 6- 10

MECHANICS OF MATERIALS Edition Fifth MECHANICS OF MATERIALS Beer • Johnston • DeWolf • Mazurek Örnek-devam _ _ Q  A. y  150x75x50  562500mm3  Q  A. y  50x225x50 c c VQG 200 x10 x585937,5 79687500x75 3  max    19,6MPa ItG _ Q  A. y  125x75x62.5  585937,5mm3 G VQG 200 x10 x585937,5 79687500x75 3  max    19,6MPa ItG © 2009 The McGraw-Hill Companies, Inc. All rights reserved.

MECHANICS OF MATERIALS Edition Fifth MECHANICS OF MATERIALS Beer • Johnston • DeWolf • Mazurek Shearing Stresses xy in Common Types of Beams İnce dikdörtgen kesitli kiriş,  2  VQ 3 V  y    Ib 2 A 1   xy  2   c  3 V  max  2 A I kesitli kirişte kayma gerilmesi dağılımı  VQ It  ort V   max Aweb © 2009 The McGraw-Hill Companies, Inc. All rights reserved.

MECHANICS OF MATERIALS Edition Fifth MECHANICS OF MATERIALS Beer • Johnston • DeWolf • Mazurek Örnek V Üç tahta parçası şekildeki gibi çivilenerek bir I profil oluşturulmuştur. Çiviler arası mesafe (kiriş ekseni boyunca) s=25 mm ve kesitteki düşey kesme kuvveti V = 500 N olduğuna göre; Her bir çivinin taşıdığı kesme kuvvetini bulunuz. Eğer çivi yerine yapıştırıcı kullanılsaydı kesitteki gerilme dağılımı nasıl olurdu? Çözüm VQ (500N)(120 106 m3) 16.20 10-6 m4 m F  (0.025m)q  (0.025m)(3704 N m q   I  3704 N Q  Ay  0.020 m  0.100 m0.060 m  120 106 m3 F  92.6 N I  1 0.020 m0.100 m3 12  2[ 1 0.100 m0.020 m3 12  0.020 m  0.100 m0.060 m2 ]  16.20 106 m4 © 2009 The McGraw-Hill Companies, Inc. All rights reserved.